You have found the following ages (in years) of 4 lions. The lions are randomly selected from the 22 lions at your local zoo: $ 1,\enspace 6,\enspace 1,\enspace 1$ Based on your sample, what is the average age of the lions? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 22 lions, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{1.69} + {13.69} + {1.69} + {1.69}} {{4 - 1}} $ $ {s^2} = \dfrac{{18.76}}{{3}} = {6.25\text{ years}^2} $ We can estimate that the average lion at the zoo is 2.3 years old. There is a variance of 6.25 years $^2$.